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L05: Query Processing & Tuning Queries Query Processing -- Principles Tuning Queries H.Lu/HKUST Query Processing & Optimization Any high-level query (SQL) on a database must be processed, optimized and executed by the DBMS The high-level query is scanned, and parsed to check for syntactic correctness An internal representation of a query is created, which is either a query tree or a query graph The DBMS then devises an execution strategy for retrieving the result of the query. (An execution strategy is a plan for executing the query by accessing the data, and storing the intermediate results) The process of choosing one out of the many execution strategies is known as query optimization H.Lu/HKUST L04: Physical Database Design (2) -- 2 Query Processor & Query Optimizer A query processor is a module in the DBMS that performs the tasks to process, to optimize, and to generate execution strategy for a high-level query Queries expressed in SQL can have multiple equivalent relational algebra query expressions. The query optimizer must select the optimal one H.Lu/HKUST L04: Physical Database Design (2) -- 3 Database Query Processing Quer y Par ser I nt ernal Represent at i on Quer Opt i mi zor Quer y Resul t s Runt i me Suppor t Dat abase H.Lu/HKUST St or age Query Execut i on Pl an Execut abl e Code Code Gener at i on St at i st i cs Dat abase L04: Physical Database Design (2) -- 4 Relational Operations We will consider how to implement: Selection : Selects a subset of rows from relation. Projection : Deletes unwanted columns from relation. Join : Allows us to combine two relations. Since each operation returns a relation, opreations can be composed! H.Lu/HKUST L04: Physical Database Design (2) -- 5 Simple Selections SELECT * FROM Reserves R WHERE R.rname < ‘C%’ Of the form R . a ttr o p v a lu e ( R ) Size of result approximated as size of R * reduction factor; we will consider how to estimate reduction factors later. With no index, unsorted: Must essentially scan the whole relation; cost is |R| (# pages in R). With an index on selection attribute: Use index to find qualifying data entries, then retrieve corresponding data records. (Hash index useful only for equality selections.) H.Lu/HKUST L04: Physical Database Design (2) -- 6 Using an Index for Selections Cost depends on #qualifying tuples, and clustering. Cost of finding qualifying data entries (typically small) plus cost of retrieving records (could be large w/o clustering). In example, assuming uniform distribution of names, about 10% of tuples qualify (100 pages, 10000 tuples). With a clustered index, cost is little more than 100 I/Os; if unclustered, upto 10000 I/Os! Important refinement for unclustered indexes: 1. Find qualifying data entries. 2. Sort the rid’s of the data records to be retrieved. 3. Fetch rids in order. This ensures that each data page is looked at just once (though # of such pages likely to be higher than with clustering). H.Lu/HKUST L04: Physical Database Design (2) -- 7 Selection - Summary R.attr op value (R) Size of result = R * selectivity Processing methods Table scan Index scan SELECT * FROM Reserves R WHERE R.rname < ‘C%’ • B+-tree, Hash index • Clustered vs. non-clustered – Clustered index: Good – Non-clustered index: • Good for low selectivity • Worse than scan for high selectivity H.Lu/HKUST L04: Physical Database Design (2) -- 8 General Selection Conditions (day<8/9/94 AND rname=‘Paul’) OR bid=5 OR sid=3 Such selection conditions are first converted to conjunctive normal form (CNF): (day<8/9/94 OR bid=5 OR sid=3 ) AND (rname=‘Paul’ OR bid=5 OR sid=3) We only discuss the case with no ORs (a conjunction of terms of the form attr op value). An index matches (a conjunction of) terms that involve only attributes in a prefix of the search key. Index on <a, b, c> matches a=5 AND b= 3, but not b=3. H.Lu/HKUST L04: Physical Database Design (2) -- 9 Two Approaches to General Selections First approach: Find the most selective access path, retrieve tuples using it, and apply any remaining terms that don’t match the index: Most selective access path: An index or file scan that we estimate will require the fewest page I/Os. Terms that match this index reduce the number of tuples retrieved; other terms are used to discard some retrieved tuples, but do not affect number of tuples/pages fetched. Consider day<8/9/94 AND bid=5 AND sid=3. A B+ tree index on day can be used; then, bid=5 and sid=3 must be checked for each retrieved tuple. Similarly, a hash index on <bid, sid> could be used; day < 8/9/94 must then be checked. H.Lu/HKUST L04: Physical Database Design (2) -- 10 Intersection of Rids Second approach (if we have 2 or more matching indexes that use Alternatives (2) or (3) for data entries): Get sets of rids of data records using each matching index. Then intersect these sets of rids (we’ll discuss intersection soon!) Retrieve the records and apply any remaining terms. Consider day<8/9/94 AND bid=5 AND sid=3. If we have a B+ tree index on day and an index on sid, both using Alternative (2), we can retrieve rids of records satisfying day<8/9/94 using the first, rids of recs satisfying sid=3 using the second, intersect, retrieve records and check bid=5. H.Lu/HKUST L04: Physical Database Design (2) -- 11 The Projection Operation SELECT DISTINCT FROM R.sid, R.bid Reserves R An approach based on sorting: Modify Pass 0 of external sort to eliminate unwanted fields. Thus, runs of about 2B pages are produced, but tuples in runs are smaller than input tuples. (Size ratio depends on # and size of fields that are dropped.) Modify merging passes to eliminate duplicates. Thus, number of result tuples smaller than input. (Difference depends on # of duplicates.) Cost: In Pass 0, read original relation (size M), write out same number of smaller tuples. In merging passes, fewer tuples written out in each pass. Using Reserves example, 1000 input pages reduced to 250 in Pass 0 if size ratio is 0.25 H.Lu/HKUST L04: Physical Database Design (2) -- 12 Projection Based on Hashing Partitioning phase: Read R using one input buffer. For each tuple, discard unwanted fields, apply hash function h1 to choose one of B-1 output buffers. Result is B-1 partitions (of tuples with no unwanted fields). 2 tuples from different partitions guaranteed to be distinct. Duplicate elimination phase: For each partition, read it and build an in-memory hash table, using hash fn h2 (<> h1) on all fields, while discarding duplicates. If partition does not fit in memory, can apply hash-based projection algorithm recursively to this partition. Cost: For partitioning, read R, write out each tuple, but with fewer fields. This is read in next phase. H.Lu/HKUST L04: Physical Database Design (2) -- 13 Discussion of Projection Sort-based approach is the standard; better handling of skew and result is sorted. If an index on the relation contains all wanted attributes in its search key, can do index-only scan. Apply projection techniques to data entries (much smaller!) If an ordered (i.e., tree) index contains all wanted attributes as prefix of search key, can do even better: Retrieve data entries in order (index-only scan), discard unwanted fields, compare adjacent tuples to check for duplicates. H.Lu/HKUST L04: Physical Database Design (2) -- 14 Equality Joins With One Join Column SELECT * FROM Reserves R1, Sailors S1 WHERE R1.sid=S1.sid In algebra: R S. Common! Must be carefully optimized. R S is large; so, R S followed by a selection is inefficient. Cost metric: # of I/Os. We will ignore output costs. H.Lu/HKUST L04: Physical Database Design (2) -- 15 Notations |R| = number of pages in outer table R ||R|| = number of tuples in outer table R |S| = number of pages in inner table S ||S|| = number of tuples in inner table S M = number of main memory pages allocated H.Lu/HKUST L04: Physical Database Design (2) -- 16 Example of Join sid 22 28 31 44 58 sname rating age dustin 7 45.0 yuppy 9 35.0 lubber 8 55.5 guppy 5 35.0 rusty 10 35.0 sid sname 31 lubber 58 rusty H.Lu/HKUST rating 8 10 sid bid day 31 101 10/11/96 58 103 11/12/96 age 55.5 35.0 rname lubber dustin SELECT * FROM Sailors R, Reserve S WHERE R.sid=S.sid bid 101 103 day rname 10/11/96 lubber 11/12/96 dustin L04: Physical Database Design (2) -- 17 Schema for Examples Sailors (sid: integer, sname: string, rating: integer, age: real) Reserves (sid: integer, bid: integer, day: dates, rname: string) Similar to old schema; rname added for variations. Reserves: Each tuple is 40 bytes long, 100 tuples per page, 1000 pages. Sailors: Each tuple is 50 bytes long, 80 tuples per page, 500 pages. H.Lu/HKUST L04: Physical Database Design (2) -- 18 Simple Nested Loops Join foreach tuple r in R do foreach tuple s in S do if ri == sj then add <r, s> to result For each tuple in the outer relation R, we scan the entire inner relation S. Cost: |R| + ||R|| * |S| = 1000 + 100*1000*500 I/Os. Page-oriented Nested Loops join: For each page of R, get each page of S, and write out matching pairs of tuples <r, s>, where r is in R-page and S is in S-page. Cost: |R| + |R|*|S| = 1000 + 1000*500 If smaller relation (S) is outer, cost = 500 + 500*1000 H.Lu/HKUST L04: Physical Database Design (2) -- 19 Block Nested Loops Join Use one page as an input buffer for scanning the inner S, one page as the output buffer, and use all remaining pages to hold ``block’’ of outer R. For each matching tuple r in R-block, s in S-page, add <r, s> to result. Then read next Rblock, scan S, etc. H.Lu/HKUST Read in block by block Hash Table for R (M-2) pages h( ) Input Buffer Output Buffer Read in page by page S R R S L04: Physical Database Design (2) -- 20 Block Nested Loop Join Scan inner table S per block of (M – 2) pages of R tuples Each scan costs |S| pages |R| / (M – 2) blocks of R tuples |R| pages for outer table R Total cost = |R| + |R| / (M – 2) * |S| pages R should be the smaller table H.Lu/HKUST L04: Physical Database Design (2) -- 21 Examples of Block Nested Loops Cost: Scan of outer + #outer blocks * scan of inner #outer blocks =no of pages of outer / blocksize With Reserves (R) as outer, and 100 pages of R: Cost of scanning R is 1000 I/Os; a total of 10 blocks. Per block of R, we scan Sailors (S); 10*500 I/Os. If space for just 90 pages of R, we would scan S 12 times. With 100-page block of Sailors as outer: Cost of scanning S is 500 I/Os; a total of 5 blocks. Per block of S, we scan Reserves; 5*1000 I/Os. With sequential reads considered, analysis changes: may be best to divide buffers evenly between R and S. H.Lu/HKUST L04: Physical Database Design (2) -- 22 Index Nested Loops Join Use one page as an input buffer for scanning the outer R, one page as the output buffer. Use all remaining pages to hold index of inner S. For each tuple r in R- page, probe index of S to find matches, s, add <r, s> to result. Then read next Rpage, repeat. H.Lu/HKUST Read in page by page Input Buffer Index for S S R Output Buffer R S L04: Physical Database Design (2) -- 23 Index Nested Loop Join Probe S index for matching S tuples per R tuple Probe hash index: 1.2 I/Os Probe B+ tree: 2-4 I/Os, plus retrieve matching S tuples: 1 I/O For ||R|| tuples |R| pages for outer table R Total cost = |R| + ||R|| * index retrieval Better than Block NL join only for small number of R tuples H.Lu/HKUST L04: Physical Database Design (2) -- 24 Examples of Index Nested Loops Hash-index (Alt. 2) on sid of Sailors (as inner): Scan Reserves: 1000 page I/Os, 100*1000 tuples. For each Reserves tuple: 1.2 I/Os to get data entry in index, plus 1 I/O to get (the exactly one) matching Sailors tuple. Total: 220,000 I/Os. Hash-index (Alt. 2) on sid of Reserves (as inner): Scan Sailors: 500 page I/Os, 80*500 tuples. For each Sailors tuple: 1.2 I/Os to find index page with data entries, plus cost of retrieving matching Reserves tuples. Assuming uniform distribution, 2.5 reservations per sailor (100,000 / 40,000). Cost of retrieving them is 1 or 2.5 I/Os depending on whether the index is clustered. H.Lu/HKUST L04: Physical Database Design (2) -- 25 Sort-Merge Join Sort R and S on the join column Scan sorted R and S to do a ``merge’’ (on join col.), and output result tuples. Advance scan of R until current R-tuple >= current S tuple, then advance scan of S until current S-tuple >= current R tuple; do this until current R tuple = current S tuple. At this point, all R tuples with same value in Ri (current R group) and all S tuples with same value in Sj (current S group) match; output <r, s> for all pairs of such tuples. Then resume scanning R and S. R is scanned once; each S group is scanned once per matching R tuple. (Multiple scans of an S group are likely to find needed pages in buffer.) H.Lu/HKUST L04: Physical Database Design (2) -- 26 External Sort ( K-Way Merge Sort) 3 4 6 2 9 4 8 7 5 6 3 1 2 Input Data Sorting 3 4 6 2 4 9 5 7 8 1 3 6 2 Sorted runs (L=3) First merge 2 3 4 4 5 6 7 8 9 1 2 3 6 Sorted runs (L=9) Second merge 1 H.Lu/HKUST 2 2 3 3 4 4 5 6 6 7 8 9 Sorted data L04: Physical Database Design (2) -- 27 Sort Merge Join External-sort R: 2 |R| * (logM-1 |R|/M + 1) Split R into |R|/M sorted runs each of size M: 2 |R| Merge up to (M – 1) runs repeatedly logM-1 |R|/M passes, each costing 2 |R| External-sort S: 2 |S| * (logM-1 |S|/M + 1) Merge matching tuples from sorted R and S: |R| + |S| Total cost = 2 |R| * (logM-1 |R|/M + 1) + 2 |S| * (logM-1 |S|/M + 1) + |R| + |S| If |R| < M*(M-1), cost = 5 * (|R| + |S|) H.Lu/HKUST L04: Physical Database Design (2) -- 28 Example of Sort-Merge Join sid 22 28 31 44 58 sname rating age dustin 7 45.0 yuppy 9 35.0 lubber 8 55.5 guppy 5 35.0 rusty 10 35.0 sid 28 28 31 31 31 58 bid 103 103 101 102 101 103 day 12/4/96 11/3/96 10/10/96 10/12/96 10/11/96 11/12/96 rname guppy yuppy dustin lubber lubber dustin Cost of merge is M+N, could be M*N (very unlikely!) With 35, 100 or 300 buffer pages, both Reserves and Sailors can be sorted in 2 passes; total join cost: 7500. (BNL cost: 2500 to 15000 I/Os) H.Lu/HKUST L04: Physical Database Design (2) -- 29 Hash Join – The Principle Ri R S n Ri Si i 1 If Ri Sj S Si for i j R H.Lu/HKUST L04: Physical Database Design (2) -- 30 Original Relation OUTPUT 1 Hash-Join Partition Phase: Partition both relations using hash fn h1: R tuples in partition i will only match S tuples in partition i. Joining Phase: Read in a partition of R, hash it using h2 (<> h1!). Scan matching partition of S, search for matches. H.Lu/HKUST Partitions 1 2 INPUT 2 hash function ... h1 M-1 M-1 Disk M main memory buffers Partitions of R & S Disk Join Result hash fn Hash table for partition Ri (k < M -1 pages) h2 h2 Input buffer for Si Disk Output buffer M main memory buffers Disk L04: Physical Database Design (2) -- 31 Hash Join Algorithm Partition phase: 2 (|R| + |S|) Partition table R using hash function h1: 2 |R| Partition table S using hash function h1: 2 |S| R tuples in partition i will match only S tuples in partition i R (M – 1) partitions, each of size |R| / (M – 1) Join phase: |R| + |S| Read in a partition of R (|R| / (M – 1) < M -1) Hash it using function h2 (<> h1!) Scan corresponding S partition, search for matches Total cost = 3 (|R| + |S|) pages Condition: M > √f|R|, f ≈ 1.2 to account for hash table H.Lu/HKUST L04: Physical Database Design (2) -- 32 General Join Conditions Equalities over several attributes (e.g., R.sid=S.sid AND R.rname=S.sname): For Index NL, build index on <sid, sname> (if S is inner); or use existing indexes on sid or sname. For Sort-Merge and Hash Join, sort/partition on combination of the two join columns. Inequality conditions (e.g., R.rname < S.sname): For Index NL, need (clustered!) B+ tree index. • Range probes on inner; # matches likely to be much higher than for equality joins. Hash Join, Sort Merge Join not applicable. Block NL quite likely to be the best join method here. H.Lu/HKUST L04: Physical Database Design (2) -- 33 Impact of Buffering If several operations are executing concurrently, estimating the number of available buffer pages is guesswork. Repeated access patterns interact with buffer replacement policy. e.g., Inner relation is scanned repeatedly in Simple Nested Loop Join. With enough buffer pages to hold inner, replacement policy does not matter. Otherwise, MRU is best, LRU is worst (sequential flooding). Does replacement policy matter for Block Nested Loops? What about Index Nested Loops? Sort-Merge Join? H.Lu/HKUST L04: Physical Database Design (2) -- 34 Summary of Join Operator Simple nested loop: |R| + ||R|| * |S| Block nested loop: |R| + |R| / (M – 2) * |S| Index nested loop: |R| + ||R|| * index retrieval Sort-merge: 2 |R| * (logM-1 |R|/M + 1) + 2 |S| * (logM-1 |S|/M + 1) + |R| + |S| Hash Join: 3 * (|R| + |S|) Condition: M > √f|R| H.Lu/HKUST L04: Physical Database Design (2) -- 35 Overview of Query Processing Statistics Parser Parsed Query High Level Query H.Lu/HKUST Cost Model Query Optimizer QEP Database Query Evaluator Query Result L04: Physical Database Design (2) -- 36 Query Optimization SELECT S.sname FROM Reserves R, Sailors S WHERE R.sid=S.sid AND R.bid=100 AND S.rating>5 sname bid=100 rating > 5 sid=sid Given: An SQL query joining n tables Dream: Map to most efficient plan Reserves Reality: Avoid rotten plans State of the art: Most optimizers follow System R’s technique Works fine up to about 10 joins H.Lu/HKUST Sailors L04: Physical Database Design (2) -- 37 Complexity of Query Optimization Many degrees of freedom Selection: scan versus (clustered, nonclustered) index Join: block nested loop, sort-merge, hash Relative order of the operators Exponential search space! A H.Lu/HKUST Heuristics Push the selections down Push the projections down Delay Cartesian products System R: Only leftD deep trees C B L04: Physical Database Design (2) -- 38 Equivalences in Relational Algebra Selection: c1... cn R c1 . . . cn R - cascade c1 c 2 R c 2 c1 R - commutative Projection: a1 R a1 . . . an R Join: R (S T) (R S) H.Lu/HKUST (RS) T (S R) - cascade - associative - commutative L04: Physical Database Design (2) -- 39 Equivalences in Relational Algebra A projection commutes with a selection that only uses attributes retained by the projection Selection between attributes of the two arguments of a cross-product converts cross-product to a join A selection on just attributes of R commutes with join R S (i.e., (R S) (R) S ) Similarly, if a projection follows a join R S, we can `push’ it by retaining only attributes of R (and S) that are needed for the join or are kept by the projection H.Lu/HKUST L04: Physical Database Design (2) -- 40 System R Optimizer 1. 2. 3. 4. 5. 6. Find all plans for accessing each base table For each table • Save cheapest unordered plan • Save cheapest plan for each interesting order • Discard all others Try all ways of joining pairs of 1-table plans; save cheapest unordered + interesting ordered plans Try all ways of joining 2-table with 1-table Combine k-table with 1-table till you have full plan tree At the top, to satisfy GROUP BY and ORDER BY • Use interesting ordered plan • Add a sort node to unordered plan H.Lu/HKUST L04: Physical Database Design (2) -- 41 Source: Selinger et al, “Access Path Selection in a Relational Database Management System” H.Lu/HKUST L04: Physical Database Design (2) -- 42 Note: Only branches for NL join are shown here. Additional branches for other join methods (e.g. sort-merge) are not shown. Source: Selinger et al, “Access Path Selection in a Relational Database Management System” H.Lu/HKUST L04: Physical Database Design (2) -- 43 What is “Cheapest”? Need information about the relations and indexes involved Catalogs typically contain at least: # tuples (NTuples) and # pages (NPages) for each relation. # distinct key values (NKeys) and NPages for each index. Index height, low/high key values (Low/High) for each tree index. Catalogs updated periodically. Updating whenever data changes is too expensive; lots of approximation anyway, so slight inconsistency ok. More detailed information (e.g., histograms of the values in some field) are sometimes stored. H.Lu/HKUST L04: Physical Database Design (2) -- 44 Estimating Result Size SELECT attribute list FROM relation list WHERE term1 AND ... AND termk Consider a query block: Maximum # tuples in result is the product of the cardinalities of relations in the FROM clause. Reduction factor (RF) associated with each termi reflects the impact of the term in reducing result size Term col=value has RF 1/NKeys(I) Term col1=col2 has RF 1/MAX(NKeys(I1), NKeys(I2)) Term col>value has RF (High(I)-value)/(High(I)-Low(I)) Result cardinality = Max # tuples * product of all RF’s. Implicit assumption that terms are independent! H.Lu/HKUST L04: Physical Database Design (2) -- 45 Cost Estimates for Single-Table Plans Index I on primary key matches selection: Cost is Height(I)+1 for a B+ tree, about 1.2 for hash index. Clustered index I matching one or more selects: (NPages(I)+NPages(R)) * product of RF’s of matching selects. Non-clustered index I matching one or more selects: (NPages(I)+NTuples(R)) * product of RF’s of matching selects. Sequential scan of file: NPages(R). Note: Typically, no duplicate elimination on projections! (Exception: Done on answers if user says DISTINCT.) H.Lu/HKUST L04: Physical Database Design (2) -- 46 Counting the Costs SELECT S.sname With 5 buffers, cost of plan: Scan Reserves (1000) + write temp FROM Reserves R, Sailors S T1 (10 pages, if we have 100 WHERE R.sid=S.sid AND boats, uniform distribution) R.bid=100 AND S.rating>5 Scan Sailors (500) + write temp T2 (250 pages, if we have 10 ratings). Sort T1 (2*10*2), sort T2 (On-the-fly) sname (2*250*4), merge (10+250), total=2300 Total: 4060 page I/Os (Sort-Merge Join) If we used BNL join, join cost = sid=sid 10+4*250, total cost = 2770 If we ‘push’ projections, T1 has only(Scan; (Scan; rating > 5 bid=100 write to write to sid, T2 only sid and sname: temp T2) temp T1) T1 fits in 3 pages, cost of BNL drops to under 250 pages, total < Reserves Sailors 2000 H.Lu/HKUST L04: Physical Database Design (2) -- 47 Exercise Reserves: 100,000 tuples, 100 tuples per page With clustered index on bid of Reserves, we get 100,000/100 = 1000 tuples on 1000/100 = 10 pages Join column sid is a key for Sailors - at most one matching tuple Decision not to push rating>5 before the join is based on availability of sid index on Sailors Cost: Selection of Reserves tuples (10 I/Os); for each tuple, must get matching Sailors tuple (1000*1.2); total 1210 I/Os H.Lu/HKUST sname (On-the-fly) rating > 5 (On-the-fly) sid=sid bid=100 (Index Nested Loops, with pipelining ) Sailors (Use hash Index on sid) (Use Reserves clustered index on sid) L04: Physical Database Design (2) -- 48 L05: Query Processing & Tuning Queries H.Lu/HKUST Query Processing – Principles Tuning Queries Avoid Redundant DISTINCT SELECT DISTINCT ssnum FROM Employee WHERE dept = ‘information systems’ DISTINCT usually entails a sort operation Slow down query optimization because one more “interesting” order to consider Remove if you know the result has no duplicates H.Lu/HKUST L04: Physical Database Design (2) -- 50 Change Nested Queries to Join SELECT ssnum FROM Employee WHERE dept IN (SELECT dept FROM Techdept) Might not use index on Employee.dept SELECT ssnum FROM Employee, Techdept WHERE Employee.dept = Techdept.dept Need DISTINCT if an employee might belong to multiple departments H.Lu/HKUST L04: Physical Database Design (2) -- 51 Avoid Unnecessary Temp Tables SELECT * INTO Temp FROM Employee WHERE salary > 40000 SELECT ssnum FROM Temp WHERE Temp.dept = ‘information systems’ Creating temp table causes update to catalog Cannot use any index on original table SELECT ssnum FROM Employee WHERE Employee.dept = ‘information systems’ AND salary > 40000 H.Lu/HKUST L04: Physical Database Design (2) -- 52 Avoid Complicated Correlation Subqueries SELECT ssnum FROM Employee e1 WHERE salary = (SELECT MAX(salary) FROM Employee e2 WHERE e2.dept = e1.dept Search all of e2 for each e1 record! SELECT MAX(salary) as bigsalary, dept INTO Temp FROM Employee GROUP BY dept SELECT ssnum FROM Employee, Temp WHERE salary = bigsalary AND Employee.dept = Temp.dept H.Lu/HKUST L04: Physical Database Design (2) -- 53 Avoid Complicated Correlation Subqueries > 1000 Throughput improvement percent 80 > 10000 70 60 50 SQLServer 2000 Oracle 8i DB2 V7.1 40 30 20 10 0 correlated subquery -10 H.Lu/HKUST SQL Server 2000 does a good job at handling the correlated subqueries (a hash join is used as opposed to a nested loop between query blocks) The techniques implemented in SQL Server 2000 are described in “Orthogonal Optimization of Subqueries and Aggregates” by C.GalindoLegaria and M.Joshi, SIGMOD 2001. L04: Physical Database Design (2) -- 54 Join on Clustering and Integer Attributes SELECT Employee.ssnum FROM Employee, Student WHERE Employee.name = Student.name Employee is clustered on ssnum ssnum is an integer SELECT Employee.ssnum FROM Employee, Student WHERE Employee.ssnum = Student.ssnum H.Lu/HKUST L04: Physical Database Design (2) -- 55 Avoid HAVING when WHERE is enough SELECT AVG(salary) as avgsalary, dept FROM Employee GROUP BY dept HAVING dept = ‘information systems’ May first perform grouping for all departments! SELECT AVG(salary) as avgsalary FROM Employee WHERE dept = ‘information systems’ GROUP BY dept H.Lu/HKUST L04: Physical Database Design (2) -- 56 Avoid Views with unnecessary Joins CREATE VIEW Techlocation AS SELECT ssnum, Techdept.dept, location FROM Employee, Techdept WHERE Employee.dept = Techdept.dept SELECT dept FROM Techlocation WHERE ssnum = 4444 Join with Techdept unnecessarily SELECT dept FROM Employee WHERE ssnum = 4444 H.Lu/HKUST L04: Physical Database Design (2) -- 57 Aggregate Maintenance Materialize an aggregate if needed “frequently” Use trigger to update create trigger updateVendorOutstanding on orders for insert as update vendorOutstanding set amount = (select vendorOutstanding.amount+sum(inserted.quantity*item.price) from inserted,item where inserted.itemnum = item.itemnum ) where vendor = (select vendor from inserted) ; H.Lu/HKUST L04: Physical Database Design (2) -- 58 Avoid External Loops No loop: sqlStmt = “select * from lineitem where l_partkey <= 200;” odbc->prepareStmt(sqlStmt); odbc->execPrepared(sqlStmt); Loop: sqlStmt = “select * from lineitem where l_partkey = ?;” odbc->prepareStmt(sqlStmt); for (int i=1; i<200; i++) { odbc->bindParameter(1, SQL_INTEGER, i); odbc->execPrepared(sqlStmt); } H.Lu/HKUST L04: Physical Database Design (2) -- 59 throughput (records/sec) Avoid External Loops 600 500 400 300 200 100 0 loop no loop SQL Server 2000 on Windows 2000 Crossing the application interface has a significant impact on performance H.Lu/HKUST Let the DBMS optimize set operations L04: Physical Database Design (2) -- 60 Avoid Cursors No cursor select * from employees; Cursor DECLARE d_cursor CURSOR FOR select * from employees; OPEN d_cursor while (@@FETCH_STATUS = 0) BEGIN FETCH NEXT from d_cursor END CLOSE d_cursor go H.Lu/HKUST L04: Physical Database Design (2) -- 61 Throughput (records/sec) Avoid Cursors 5000 4000 3000 2000 1000 0 cursor H.Lu/HKUST SQL SQL Server 2000 on Windows 2000 Response time is a few seconds with a SQL query and more than an hour iterating over a cursor L04: Physical Database Design (2) -- 62 Retrieve Needed Columns Only H.Lu/HKUST Throughput (queries/msec) All Select * from lineitem; Covered subset Select l_orderkey, l_partkey, l_suppkey, l_shipdate, l_commitdate from lineitem; Avoid transferring unnecessary data May enable use of a covering index. 1.75 1.5 1.25 all covered subset 1 0.75 0.5 0.25 0 no index index L04: Physical Database Design (2) -- 63 Use Direct Path for Bulk Loading sqlldr directpath=true control=load_lineitem.ctl data=E:\Data\lineitem.tbl load data infile "lineitem.tbl" into table LINEITEM append fields terminated by '|' ( L_ORDERKEY, L_PARTKEY, L_SUPPKEY, L_LINENUMBER, L_QUANTITY, L_EXTENDEDPRICE, L_DISCOUNT, L_TAX, L_RETURNFLAG, L_LINESTATUS, L_SHIPDATE DATE "YYYYMM-DD", L_COMMITDATE DATE "YYYY-MM-DD", L_RECEIPTDATE DATE "YYYY-MM-DD", L_SHIPINSTRUCT, L_SHIPMODE, L_COMMENT ) H.Lu/HKUST L04: Physical Database Design (2) -- 64 Use Direct Path for Bulk Loading Throughput (rec/sec) 50000 40000 30000 20000 10000 65 0 conventional H.Lu/HKUST direct path insert Direct path loading bypasses the query engine and the storage manager. It is orders of magnitude faster than for conventional bulk load (commit every 100 records) and inserts (commit for each record). L04: Physical Database Design (2) -- 65 Some Idiosyncrasies OR may stop the index being used break the query and use UNION Order of tables may affect join implementation H.Lu/HKUST L04: Physical Database Design (2) -- 66 Query Tuning Principles Avoid redundant DISTINCT Change nested queries to join Avoid unnecessary temp tables Avoid complicated correlation subqueries Join on clustering and integer attributes Avoid HAVING when WHERE is enough Avoid views with unnecessary joins Maintain frequently used aggregates Avoid external loops H.Lu/HKUST L04: Physical Database Design (2) -- 67 Query Tuning Principles Avoid cursors Retrieve needed columns only Use direct path for bulk loading H.Lu/HKUST L04: Physical Database Design (2) -- 68